Question

The line 3y=4x -15 intersects the curve 8x2 = 45 + 27y2 at the points A & B. Find the coordinates of A and B.

Answer 1

SOLUTION

GIVEN

The line 3y = 4x - 15 intersects the curve 8x² = 45 + 27y² at the points A & B.

TO DETERMINE

The coordinates of A and B

EVALUATION

Here the given equation of the curve is

8x² = 45 + 27y² - - - - - - (1)

The given equation of the line is

3y = 4x - 15 - - - - (2)

For point of intersection we have

8x² = 45 + 3 × 9y²

⇒ 8x² = 45 + 3 × ( 4x - 15)²

⇒ 8x² = 45 + 48x² - 360x + 675

⇒ 40x² - 360x + 720 = 0

⇒ x² - 9x + 18 = 0

⇒ (x - 3)(x - 6) = 0

Now x - 3 = 0 gives x = 3

x - 6 = 0 gives x = 6

For x = 3 we get y = - 1

For x = 6 we have y = 3

So the points of intersections are

A (3, - 1) & B( 6,3)

FINAL ANSWER

Hence the required points are

A (3, - 1) & B ( 6,3)

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Answer 2

Step-by-step explanation:

Given:The line 3y=4x -15 intersects the curve 8x² = 45 + 27y² at the points A & B.

To find: Find the coordinates of A and B.

Solution:

Step 1: Put the value of 3y from line into the curve.

$8 {x}^{2} = 45 + 3 \times 9 {y}^{2} \\ \\ or \\ \\ 8 {x}^{2} = 45 + 3 \times {(3y)}^{2} \\ \\ 8 {x}^{2} = 45 + 3 \times {(4x - 15)}^{2}$

Step 2: Solve the equation for x

$8 {x}^{2} = 45 + 3 \times (16 {x}^{2} - 120x + 225) \\ \\ 8 {x}^{2} = 45 + 48 {x}^{2} - 360x + 675 \\ \\ - 40 {x}^{2} + 360x - 720 = 0 \\ \\ or \\ \\ {x}^{2} - 9x + 18 = 0 \\ \\ or \\ \\ {x}^{2} - 6x - 3x + 18 = 0 \\ \\ x(x - 6) - 3(x - 6) = 0 \\ \\ (x - 6)(x - 3) = 0 \\ \\ x = 6 \\ \\ or \\ \\ x = 3$

Step 3: Put the values of x in line and find the values of y.

When x=6

$3y = 4 \times 6 - 15 \\ \\ 3y = 24 - 15 \\ \\ 3y = 9 \\ \\ y = \frac{9}{3} \\ \\ y = 3$

Coordinates of A(6,3)

When x=3

$3y = 4 \times 3 - 15 \\ \\ 3y = 12 - 15 \\ \\ 3y = - 3 \\ \\ y = - 1$

Coordinates of B (3,-1).

Final answer:

Coordinates of A and B are (6,3) and (3,-1).

Hope it helps you.

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